What is the extraneous solution to these equations? $\dfrac{x^2 - 43}{x - 4} = \dfrac{-16x - 107}{x - 4}$
Answer: Multiply both sides by $x - 4$ $ \dfrac{x^2 - 43}{x - 4} (x - 4) = \dfrac{-16x - 107}{x - 4} (x - 4)$ $ x^2 - 43 = -16x - 107$ Subtract $-16x - 107$ from both sides: $ x^2 - 43 - (-16x - 107) = -16x - 107 - (-16x - 107)$ $ x^2 - 43 + 16x + 107 = 0$ $ x^2 + 64 + 16x = 0$ Factor the expression: $ (x + 8)(x + 8) = 0$ Therefore $x = -8$ The original expression is defined at $x = -8$ and $x = -8$, so there are no extraneous solutions.